So in the class, Luis solved this problem by dealing with each new person as a new independent event, and for each one, the sample space of days to be “on a different day than all the others” is decremented. But how if I wanted to solve this problem by actually considering the probability for each new person of having the same birthday as on of the others, how could I go about doing that?
Hi @Osama_Adel!
Well, tackling this problem directly like this can be quite challenging because it involves several possibilities for each draw. For each draw, you would have to compute the the probability that it is one of the previous and you must consider all the cases: the case where the draw matches only 1 birthday, 2 birthdays, … n-1 birthdays (it might be the case where several students have the same birthday).
Note that “on a different day than all the others” = 1 - “at least one is equal to the others”
And “at least one is equal to the others” = exactly one is equal to the others + exactly two are equal to the others + … + exactly n-1 is equal to the others.
So, let’s just use the theory in our favour to simplify the calculations! 
Was that clear?
Cheers,
Lucas